24th Putnam 1963

Let D be the differential operator d/dx and E the differential operator xD(xD - 1)(xD - 2) ... (xD - n). Find an expression of the form y = ∫₁^x g(t) dt for the solution of the differential equation Ey = f(x), with initial conditions y(1) = y'(1) = ... = y⁽ⁿ⁾(1) = 0, where f(x)is a continuous real-valued function on the reals.

Solution

The first step is to try to simplify Ey. Experimenting with small n, we soon suspect that Ey = xⁿ⁺¹y⁽ⁿ⁺¹⁾ and that is easily proved by induction.

So we seek a solution to xⁿ⁺¹y⁽ⁿ⁺¹⁾ = f(x), subject to y(1) = y'(1) = ... = y⁽ⁿ⁾(1) = 0. The trick is to use the Taylor expansion: y(x) = y(1) + (x-1)y'(1) + (x-1)²/2! y''(1) + ... + (x-1)ⁿy⁽ⁿ⁾(1) + ∫₁^x g(t) dt, where g(t) = (x - t)ⁿyⁿ⁺¹(t)/n! = (x - t)ⁿf(t)/(n! tⁿ⁺¹).

24th Putnam 1963

© John Scholes
jscholes@kalva.demon.co.uk
5 Feb 2002